Pigeonhole Principle - types of Universes

[falkor2k15]

I've been experimenting with this for over a year now. The same mathematical laws about PHP apply across the board to all collection of spins - now on referred to as "Universes". However, some things are subject to change, which I believe is how we can exploit loopholes in Roulette. Can anyone add to this  list - "set in stone" and "subject to change"? I prefer to use dozens as an example (excluding zero).

DOZEN REPEATS

Set in stone

1) A dozen has to repeat in 4 spins (the basis for Non-Random). The 4 spins are the pigeons, and they have only 3 pigeonholes, so 1 pigeonhole has to contain 2 dozens = a repeat.
2) A dozen has to repeat 4 times in 13 spins. The most extreme scenario is that one pigeonhole/partition contains 5 dozens and the rest contain 4 - or 1 pigeonhole contains 5 dozens and the rest are empty.
3) The maximum number of dozens/pigeons is more than the average +1. The average Cycle Length is 2, so the maximum number of dozens has to be equal or greater than 2+1 (in this case is 3).

Subject to change

1) The pigeons and the pigeonholes have the same value in terms of uniques. Obviously, 4 spins are 4 pigeons, but there are only 3 kinds of "fruits" so to speak. We have 3 dozens and there are an equal number of partitions they can fall in before a repeat comes. So, by default it's 2/2 for Red and Black, 3/3 for dozens and 6/6 for Lines. But this is subject to change when creating your own pigeonholes and rules.
2) The pigeon values are taken from a 1 spin result. Each dozen is taken from 1 spin only.
3) The pigeons in terms of fruit are equally-likely. Each flavor has 33% chance.
4) The repeat has more chance to be the same as the starting partition = order 1. If the first dozen is 1 then it's 63% chance that the repeat will be dozen 1, i.e. 121 or 11 or 1231.
5) The dozens have more chance of repeating as position/distance = 1. There's a 63% chance of 11 or 122 or 1233, etc. In other words, the repeat has a 63% chance to end as position 1. Something similar could be said for clockwise/CC.
6) The average Cycle Length for dozens is 2.
7) Each universe of pigeons/fruits can be dependent on another in some way. The Dozens depend on High/Low and the Lines, as well as  most of the other groups, in various ways.

Next, I will show for the first 6 how they can be changed, based on creating our own Non-Random game.

[falkor2k15]

1) The pigeons and the pigeonholes have the same value in terms of uniques. Obviously, 4 spins are 4 pigeons, but there are only 3 kinds of "fruits" so to speak. We have 3 dozens and there are an equal number of partitions they can fall in before a repeat comes. So, by default it's 2/2 for Red and Black, 3/3 for dozens and 6/6 for Lines. But this is subject to change when creating your own pigeonholes and rules.

Dozens are 3/3. Below I've put all 37 numbers into 18 pigeonholes (37/18) based on a rule. Pigeonhole 7 can contain numbers 6,7,24,25, so a repeat might be number 6 followed by number 24.

Can that be exploited based on what is set in stone or subject to change?

If we could change it from 37/18 to 18/37 I reckon we might have more chance of finding an exploit. Numbers are somewhat like this naturally since although they are 37/37, they are really 37/25, but there's too many pigeons/holes here to exploit due to the table limits.

[falkor2k15]

2) The pigeon values are taken from a 1 spin result. Each dozen is taken from 1 spin only.

Here I've created 24 Number Roulette using Quads (1-9, 10-18, 19-27, 28-36) and Lines. The number of fruits are 24 and the number of pigeonholes are 24 (24/24), but each pigeon is based on 2 spins.


Spin 1 we take the Quad value and Spin 2 we take the Line value. Depending on this 2-spin combination trial we get 24 new numbers.

Quad 1 & Line 1 = New Number 1
Quad 1 & Line 2 = New Number 2

How to bet for a repeat on both? Well, we bet 2 units on Quad 1, and if we win then we parlay the winnings onto Lines 1 and 2. The total payout for the 2 spins would be equivalent to 24 Number Roulette (1:23?). The only problem is that the house edge would increase based on zero appearing during spin 1 or spin 2 of our trial.

3 Quad 1
6 Line 1 7 1 
19 Quad 3
21 Line 4 7 1 16 
31 Quad 4
29 Line 5 7 1 16 23 
30 Quad 4
5 Line 1 7 1 16 23 19 
31 Quad 4
35 Line 6 7 1 16 23 19 24 
2 Quad 1
30 Line 5 7 1 16 23 19 24 5 
20 Quad 3
24 Line 4 7 1 16 23 19 24 5 16  NCL7o3

24 Quad 3
11 Line 2 16 14 
11 Quad 2
15 Line 3 16 14 9 
24 Quad 3
19 Line 4 16 14 9 16  NCL3o1

10 Quad 2
8 Line 2 16 8 
17 Quad 2
12 Line 2 16 8 8  NCL2o2

Can that be exploited based on what is set in stone or subject to change?

[falkor2k15]

3) The pigeons in terms of fruit are equally-likely. Each flavor has 33% chance.

Here we take the Cycle Length and Order from a dozen cycle to create 6 new fruits and pigeonholes (Options Cycle).

Each pigeon is now not equally-likely:
Opt. 1: CL1o1   33%
Opt. 2: CL2o1   22%
Opt. 3: CL2o2   22%
Opt. 4: CL3o1   7%
Opt. 5: CL3o2   7%
Opt. 6: CL3o3   7%



Can that be exploited based on what is set in stone or subject to change?

[RouletteGhost]

What are you looking to achieve?

Me thinks that no matter what you do the HE remains.

[falkor2k15]

4) The repeat has more chance to be the same as the starting partition = order 1. If the first dozen is 1 then it's 63% chance that the repeat will be dozen 1, i.e. 121 or 11 or 1231.
5) The dozens have more chance of repeating as position/distance = 1. There's a 63% chance of 11 or 122 or 1233, etc. In other words, the repeat has a 63% chance to end as position 1. Something similar could be said for clockwise/CC.

Does the above change for the previous 6 Dozen Options example?

6) The average Cycle Length for dozens is 2.

Of course this changes depending in the type of cycle: sometimes CL1 is more common - other times it's CL4+, such as on lines or streets.

Looking at the above can we find an exploit/loophole based on what is set in stone or subject to change?

[falkor2k15]

1) The pigeons and the pigeonholes have the same value in terms of uniques. Obviously, 4 spins are 4 pigeons, but there are only 3 kinds of "fruits" so to speak. We have 3 dozens and there are an equal number of partitions they can fall in before a repeat comes. So, by default it's 2/2 for Red and Black, 3/3 for dozens and 6/6 for Lines. But this is subject to change when creating your own pigeonholes and rules.

We can also have dynamically changing pigeonholes too:


Normally, Lines go up to Cycle Length 6 maximum, but in cases like above we can interchange between the standard 6 lines and 5 in-between, resulting in a maximum of Cycle Length 11.

Using this fact, can you find an exploit/loophole based on what is set in stone or subject to change?

[falkor2k15]

3) The pigeons in terms of fruit are equally-likely. Each flavor has 33% chance.

7) Each universe of pigeons/fruits can be dependent on another in some way. The Dozens depend on High/Low and the Lines, as well as  most of the other groups, in various ways.

Take just the Cycle Length result (with 6 Options we took both CL and Order) from a Dozen Cycle and create new pigeons/fruits/pigeonholes from it:
CL1: 33%
CL2: 44%
CL3: 22%

Here a Cycle Length has to repeat, ie. out of 3 pigeonholes 1 must contain 2 pigeons, and each pigeon is not equally-likely.

The Cycle Lengths stream is dependent on the Dozens streams.

Can you use the above to find an exploit/loophole based on what is set in stone and what is subject to change?

[falkor2k15]

4) The repeat has more chance to be the same as the starting partition = order 1. If the first dozen is 1 then it's 63% chance that the repeat will be dozen 1, i.e. 121 or 11 or 1231.
5) The dozens have more chance of repeating as position/distance = 1. There's a 63% chance of 11 or 122 or 1233, etc. In other words, the repeat has a 63% chance to end as position 1. Something similar could be said for clockwise/CC.
7) Each universe of pigeons/fruits can be dependent on another in some way. The Dozens depend on High/Low and the Lines, as well as  most of the other groups, in various ways.

Can you use both order and position in parallel to find an exploit/loophole based on what is set in stone and what is subject to change?

[RouletteGhost]

What are you looking to achieve?

Me thinks that no matter what you do the HE remains

[falkor2k15]

What are you looking to achieve?

Me thinks that no matter what you do the HE remains

Gain edge!
If you try to bet for line 1 following 10 appearances of line 2 there's no guarantee when line 1 will appear, but if you try to bet for a repeat of a line then it's guaranteed in 7 spins.
If you try to bet for a repeat of a line you may encounter a deadlock situation where all 6 lines have hit once, so we are no longer able to cover all single hits and still gain a profit.
If you try to bet for a repeat of a number you may reach the table limits if the repeat takes too long, but you will never encounter a deadlock.
If you could find a successful way to bet for a repeat then you can overcome the house edge and gain edge for the player. This is achieved by making one or more of the changes above to your pigeonhole framework that underlies the repeats mechanism. Other methods have been claimed to work too, such as hedging bets or avoiding variance - said to also work in combination with PHP.